Kết quả giới hạn Lim 1 + 3x / căn {2{x^2} + 3} là

Chọn B

\[\mathop {\lim }\limits_{x \to - \infty } \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {3 + \frac{1}{x}} \right)}}{{\left| x \right|.\sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {3 + \frac{1}{x}} \right)}}{{ - x.\sqrt {2 + \frac{3}{{{x^2}}}} }}\]

\[ = \mathop {\lim }\limits_{x \to - \infty } \frac{{3 + \frac{1}{x}}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} = \frac{{ - 3}}{{\sqrt 2 }}\].